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Article

Positive Solutions for a Semipositone Singular ψ–Riemann–Liouville Fractional Boundary Value Problem

1
Department of Computer Science and Engineering, Gh. Asachi Technical University, 700050 Iasi, Romania
2
Department of Mathematics, Gh. Asachi Technical University, 700506 Iasi, Romania
*
Author to whom correspondence should be addressed.
Mathematics 2025, 13(20), 3292; https://doi.org/10.3390/math13203292
Submission received: 15 September 2025 / Revised: 11 October 2025 / Accepted: 13 October 2025 / Published: 15 October 2025

Abstract

We explore the existence of positive solutions to a ψ –Riemann–Liouville fractional differential equation with a parameter and a sign-changing singular nonlinearity, supplemented with nonlocal boundary conditions which contain Riemann–Stieltjes integrals and various fractional derivatives. To establish our main results, we use the Guo–Krasnosel’skii fixed-point theorem.

1. Introduction

We consider the fractional differential equation
D a + α , ψ u ( t ) + λ f ( t , u ( t ) ) = 0 , t ( a , b ) ,
subject to the nonlocal boundary conditions
u ( i ) ( a ) = 0 , i = 0 , 1 , , n 2 ; D a + ς , ψ u ( b ) = i = 1 m a b D a + ϱ i , ψ u ( s ) d H i ( s ) ,
where 0 a < b , α > 0 , α ( n 1 , n ] , n N , n 3 , ψ C n [ a , b ] with ψ ( t ) > 0 for all t [ a , b ] , D a + k , ψ u denotes the ψ –Riemann–Liouville fractional derivative of function u of order k, for k { α , ς , ϱ i , i = 1 , , m } , m N , 1 ς < α 1 , 0 ϱ i ς , i = 1 , , m , λ is a positive parameter, the function f may change sign and may be singular at the points t = a and/or t = b , and H i : [ a , b ] R , i = 1 , , m are bounded variation functions.
In this paper we give intervals for the parameter λ such that the problem (1)–(2) has at least one positive solution. Since f can have negative values, our problem is called a semipositone fractional boundary value problem. In the proofs of the main results we apply the Guo–Krasnosel’skii fixed-point theorem. The ψ –Riemann–Liouville fractional derivative generalizes the Riemann–Liouville derivative (for ψ ( t ) = t ) , and the Hadamard derivative (for ψ ( t ) = ln t ) (see the definitions in Section 2). In the following, we present some papers that investigate various fractional differential equations and systems which are connected to our problem. In [1], the authors studied the nonlinear fractional differential equation
D 0 + α u ( t ) + λ f ( t , u ( t ) ) = 0 , t ( 0 , 1 ) ,
with the multi-point boundary conditions
u ( 0 ) = u ( 0 ) = = u ( n 2 ) ( 0 ) = 0 , D 0 + p u ( 1 ) = i = 1 m a i D 0 + q u ( ξ i ) ,
where α R , α ( n 1 , n ] , n N , n 3 , m N , 0 < ξ 1 < < ξ m < 1 , p , q R , 1 p n 2 , q [ 0 , p ] , λ is a positive parameter, the function f may change sign and may be singular at t = 0 and/or t = 1 , and D 0 + κ denotes the Riemann–Liouville derivative of order κ , for κ { α , p , q } . With the aid of the Guo–Krasnosel’skii fixed-point theorem, they proved the existence of positive solutions for problem (3)–(4). In [2], the author investigated the fractional differential equation
D 0 + α u ( t ) + f ( t , u ( t ) ) = 0 , t ( 0 , 1 ) ,
supplemented with the integral boundary conditions
u ( 0 ) = u ( 0 ) = = u ( n 2 ) ( 0 ) = 0 , D 0 + p u ( 1 ) = 0 1 D 0 + q u ( t ) d H ( t ) ,
where n N , n 3 , α R , α ( n 1 , n ] , p , q R , 1 p n 2 , 0 q p , the function f may change sign and may be singular at the points t = 0 , t = 1 , and/or u = 0 , and H is a bounded variation function. By way of various height functions of the nonlinearity of Equation (5) defined on special bounded sets, and the Guo–Krasnosel’skii fixed-point theorem, they proved the existence and multiplicity of positive solutions of problem (5)–(6). In [3], the authors examined the existence of multiple positive solutions for the fractional differential Equation (5) subject to the boundary conditions
u ( 0 ) = u ( 0 ) = = u ( n 2 ) ( 0 ) = 0 , D 0 + β 0 u ( 1 ) = i = 1 m 0 1 D 0 + β i u ( t ) d H i ( t ) ,
where n , m N , n 3 , α R , α ( n 1 , n ] , β i R for all i = 0 , , m , 1 β 0 < α 1 , 0 β 1 < β 2 < < β m β 0 , the function f may change sign and may be singular at the points t = 0 , t = 1 , and/or u = 0 , and H i , i = 1 , m are bounded variation functions. For the proof of the main results, they used varied height functions of f defined on special bounded sets, and two theorems from the fixed-point index theory. We also mention the papers [4,5,6,7,8,9,10], where the authors explored the existence of positive solutions for various semipositone or singular Riemann–Liouville fractional differential equations with different boundary conditions. In [11], the authors studied the existence of at least three positive solutions for a class of Riemann–Liouville fractional differential equations with p-Laplacian operator and singular sign-changing nonlinearities, supplemented with integral boundary conditions containing a positive parameter. In [12], by using the Guo–Krasnosel’skii and Leggett–Williams fixed-point index theorems, the authors analyzed the existence of positive solutions for a singular Riemann–Liouville fractional differential equation with both generalized Laplacian and positive parameter subject to nonlocal boundary conditions. In [13], the authors examined the system of nonlinear fractional differential equations
D 0 + α u ( t ) + λ f ( t , u ( t ) , v ( t ) ) = 0 , t ( 0 , 1 ) , D 0 + β v ( t ) + μ g ( t , u ( t ) , v ( t ) ) = 0 , t ( 0 , 1 ) ,
with the coupled integral boundary conditions
u ( 0 ) = u ( 0 ) = = u ( n 2 ) ( 0 ) = 0 , D 0 + p u ( 1 ) = 0 1 v ( s ) d H ( s ) , v ( 0 ) = v ( 0 ) = = v ( m 2 ) ( 0 ) = 0 , D 0 + q v ( 1 ) = 0 1 u ( s ) d K ( s ) ,
where n , m N , n , m 3 , α , β R , α ( n 1 , n ] , β ( m 1 , m ] , p , q N , 1 p n 2 , 1 q m 2 , λ and μ are positive parameters, the functions f and g are sign-changing continuous functions which may be singular at t = 0 and/or t = 1 , and H and K are bounded variation functions. They gave intervals for the parameters λ and μ such that the problem (7)–(8) has at least one positive solution. Other systems of Riemann–Liouville fractional differential equations supplemented with varied nonlocal boundary conditions were explored in the papers [14,15,16,17,18]. For detailed studies of various Riemann–Liouville, Caputo and Hadamard fractional differential equations and systems with various applications, we refer the reader to the monographs [19,20,21,22]. Applications of fractional calculus in physics and engineering can be found in the papers [23,24,25,26]. ψ –Caputo fractional differential equations or inclusions with initial or boundary conditions were studied in the papers [27,28,29,30,31]. The generalizations of ψ –Riemann–Liouville and ψ –Caputo fractional derivatives, namely ψ –Hilfer fractional derivatives with their properties were introduced in [32]. Another generalization of the ψ –Hilfer fractional derivative, that is, the ( k , ψ ) –Hilfer fractional derivative, was defined in [33]. Some fractional boundary value problems containing ( k , ψ ) –Hilfer fractional derivatives were investigated in [34,35]. Discrete fractional boundary value problems were studied in the papers [36,37,38,39,40] (with nabla discrete fractional differences), [41,42,43,44] (with delta discrete fractional differences), and [45,46] (with delta-nabla discrete fractional differences). The novelty of our paper consists in the presence of the ψ –Riemann–Liouville fractional derivatives in Equation (1) and the boundary conditions (2). In addition, the nonlinearity f from (1) is not a non-negative function, but it may change sign and may be singular at some points. Lastly, the condition from (2) with Riemann–Stieltjes integrals is a general condition containing multi-point boundary conditions (when H i , i = 1 , , m are step functions), classical integral conditions (when H i , i = 1 , , m are differentiable functions), and combinations of them.
The paper is organized as follows. Section 2 introduces the definitions and key properties of ψ –Riemann–Liouville fractional integrals and derivatives. We also examine the linear counterpart of problem (1)–(2), together with the associated Green function and its properties. Section 3 is devoted to the main existence theorems for positive solutions of problem (1)–(2). In Section 4, we provide two illustrative examples that demonstrate the applicability of our results, while Section 5 concludes this paper.

2. Auxiliary Results

In this section we will firstly present the definitions for the ψ –Riemann–Liouville fractional integral and derivative and some of their properties.
Let 0 a < b < , α > 0 , α ( n 1 , n ] , and ψ C n [ a , b ] with ψ ( t ) > 0 for all t [ a , b ] .
Definition 1
([47]-pag.99–100; [48]-pag.325). The (left-sided) Riemann–Liouville fractional integral of function u : [ a , b ] R with respect to function ψ on [ a , b ] of order α > 0 , or the (left-sided) ψ–Riemann–Liouville fractional integral of function u : [ a , b ] R of order α > 0 , is defined by
I a + α , ψ u ( t ) = 1 Γ ( α ) a t ψ ( s ) ( ψ ( t ) ψ ( s ) ) α 1 u ( s ) d s , t > a ; I a + α , ψ u ( a ) = 0 .
Definition 2
([47]-pag.101–102, [48]-pag.326). The (left-sided) Riemann–Liouville fractional derivative of function u : [ a , b ] R with respect to function ψ on [ a , b ] of order α > 0 , or the (left-sided) ψ–Riemann–Liouville fractional derivative of function u : [ a , b ] R of order α > 0 , is defined by
D a + α , ψ u ( t ) = 1 ψ ( t ) d d t n I a + n α , ψ u ( t ) = 1 ψ ( t ) d d t n 1 Γ ( n α ) a t ψ ( s ) ( ψ ( t ) ψ ( s ) ) n α 1 u ( s ) d s , t > a ,
where n = α + 1 .
The notation α stands for the largest integer not greater than α . For u C m [ a , b ] , if α = m N then D a + m , ψ u ( t ) = 1 ψ ( t ) d d t m u ( t ) , t > a , and if α = 0 then D a + 0 , ψ u ( t ) = u ( t ) for t > 0 .
We observe that the ψ –Riemann–Liouville integral and derivative are generalizations of various fractional integrals and derivatives; indeed:
-
If ψ ( t ) = t , t [ a , b ] , a 0 , then we obtain the Riemann–Liouville fractional integral/derivative of order α ;
-
If ψ ( t ) = ln t , t [ a , b ] , a > 0 , then we get the Hadamard fractional integral/derivative of order α ;
-
If ψ ( t ) = t ϱ ϱ , t [ a , b ] , a > 0 , with ϱ > 0 , then we find the Katugampola fractional integral/derivative of order α ; and so on.
Lemma 1
([47]-Lemma 2.26; [48]-pag.326). Let α , β > 0 . Then for a function u : [ a , b ] R , u L p ( a , b ) , 1 p , we have
( I a + α , ψ I a + β , ψ u ) ( t ) = ( I a + α + β , ψ u ) ( t ) , f o r a . e . t ( a , b ) .
Lemma 2.
Let α > β > 0 . Then for u L p ( a , b ) , 1 p , we have
( D a + β , ψ I a + α , ψ u ) ( t ) = ( I a + α β , ψ u ) ( t ) , f o r a . e . t ( a , b ) .
The proof of Lemma 2 is similar to the proof of Property 2.2 from [47].
Lemma 3
([47]-pag.100, 103; [48]-pag.326). Let α , β > 0 . Then, we have
(a) ( I a + α , ψ ( [ ψ ( x ) ψ ( a ) ] β 1 ) ) ( t ) = Γ ( β ) Γ ( α + β ) [ ψ ( t ) ψ ( a ) ] α + β 1 , t > a ;
(b) D a + α , ψ ( [ ψ ( x ) ψ ( a ) ] β 1 ) ( t ) = Γ ( β ) Γ ( β α ) [ ψ ( t ) ψ ( a ) ] β α 1 , t > a ;
(c) D a + α , ψ ( [ ψ ( x ) ψ ( a ) ] α j ) ( t ) = 0 , for j = 1 , 2 , , p , where p = n if α N with α ( n 1 , n ) , and p = n if α = n N .
Lemma 4.
Let α > 0 . Then for f L p ( a , b ) , 1 p ; we have
D a + α , ψ I a + α , ψ f ( t ) = f ( t ) , f o r a . e . t ( a , b ) .
Proof. 
We obtain
D a + α , ψ I a + α , ψ f ( t ) = 1 ψ ( t ) d d t n I a + n α , ψ I a + α , ψ f ( t ) = 1 ψ ( t ) d d t n I a + n , ψ f ( t ) = 1 ψ ( t ) d d t n 1 Γ ( n ) a t ψ ( s ) ( ψ ( t ) ψ ( s ) ) n 1 f ( s ) d s = 1 ψ ( t ) d d t n 1 1 ψ ( t ) d d t 1 Γ ( n ) a t ψ ( s ) ( ψ ( t ) ψ ( s ) ) n 1 f ( s ) d s = 1 ψ ( t ) d d t n 1 1 ψ ( t ) 1 Γ ( n ) a t ψ ( s ) ( n 1 ) ( ψ ( t ) ψ ( s ) ) n 2 ψ ( t ) f ( s ) d s + ψ ( t ) ( ψ ( t ) ψ ( t ) ) n 1 f ( t ) = 1 ψ ( t ) d d t n 1 1 ( n 2 ) ! a t ψ ( s ) ( ψ ( t ) ψ ( s ) ) n 2 f ( s ) d s = 1 ψ ( t ) d d t n 2 1 ψ ( t ) d d t 1 ( n 2 ) ! a t ψ ( s ) ( ψ ( t ) ψ ( s ) ) n 2 f ( s ) d s = 1 ψ ( t ) d d t n 2 1 ψ ( t ) 1 ( n 2 ) ! a t ψ ( s ) ( n 2 ) ( ψ ( t ) ψ ( s ) ) n 3 ψ ( t ) f ( s ) d s + ψ ( t ) ( ψ ( t ) ψ ( t ) ) n 2 f ( t ) = 1 ψ ( t ) d d t n 2 1 ( n 3 ) ! a t ψ ( s ) ( ψ ( t ) ψ ( s ) ) n 3 f ( s ) d s = = 1 ψ ( t ) d d t 2 1 1 ! a t ψ ( s ) ( ψ ( t ) ψ ( s ) ) f ( s ) d s = 1 ψ ( t ) d d t 1 ψ ( t ) d d t a t ψ ( s ) ( ψ ( t ) ψ ( s ) ) f ( s ) d s = 1 ψ ( t ) d d t 1 ψ ( t ) a t ψ ( s ) ψ ( t ) f ( s ) d s + ψ ( t ) ( ψ ( t ) ψ ( t ) ) f ( t ) = 1 ψ ( t ) d d t a t ψ ( s ) f ( s ) d s = 1 ψ ( t ) ψ ( t ) f ( t ) = f ( t ) , f o r a . e . t ( a , b ) .
Lemma 5.
Let α > 0 , n = α + 1 if α N , and n = α if α N . If f L 1 ( a , b ) and I a + n α , ψ f A C n [ a , b ] , then
I a + α , ψ D a + α , ψ f ( t ) = f ( t ) j = 1 n D n j ( I a + n α , ψ f ) ( a ) Γ ( α j + 1 ) ( ψ ( t ) ψ ( a ) ) α j , f o r a . e . t ( a , b ) ,
where D = 1 ψ ( t ) d d t .
Proof. 
We find
I a + α , ψ D a + α , ψ f ( t ) = 1 Γ ( α ) a t ψ ( s ) ( ψ ( t ) ψ ( s ) ) α 1 D a + α , ψ f ( s ) d s = 1 ψ ( t ) d A d t ( t ) ,
where
A ( t ) = 1 Γ ( α + 1 ) a t ψ ( s ) ( ψ ( t ) ψ ( s ) ) α D a + α , ψ f ( s ) d s .
For A ( t ) we obtain
A ( t ) = 1 Γ ( α + 1 ) a t ψ ( s ) ( ψ ( t ) ψ ( s ) ) α 1 ψ ( s ) d d s n ( I a + n α , ψ f ( s ) ) d s = 1 Γ ( α + 1 ) a t ψ ( s ) ( ψ ( t ) ψ ( s ) ) α 1 ψ ( s ) d d s 1 ψ ( s ) d d s n 1 ( I a + n α , ψ f ( s ) ) d s = 1 Γ ( α + 1 ) a t ( ψ ( t ) ψ ( s ) ) α d d s 1 ψ ( s ) d d s n 1 ( I a + n α , ψ f ( s ) ) d s = 1 Γ ( α + 1 ) ( ψ ( t ) ψ ( s ) ) α 1 ψ ( s ) d d s n 1 ( I a + n α , ψ f ( s ) ) | s = a s = t a t α ( ψ ( t ) ψ ( s ) ) α 1 ( ψ ( s ) ) 1 ψ ( s ) d d s n 1 ( I a + n α , ψ f ( s ) ) d s = 1 Γ ( α + 1 ) 0 ( ψ ( t ) ψ ( a ) ) α 1 ψ ( s ) d d s n 1 ( I a + n α , ψ f ( s ) ) | s = a + α a t ( ψ ( t ) ψ ( s ) ) α 1 ψ ( s ) 1 ψ ( s ) d d s n 1 ( I a + n α , ψ f ( s ) ) d s = 1 Γ ( α + 1 ) ( ψ ( t ) ψ ( a ) ) α 1 ψ ( s ) d d s n 1 ( I a + n α , ψ f ( s ) ) | s = a + 1 Γ ( α ) a t ψ ( s ) ( ψ ( t ) ψ ( s ) ) α 1 1 ψ ( s ) d d s n 1 ( I a + n α , ψ f ( s ) ) d s = 1 Γ ( α + 1 ) ( ψ ( t ) ψ ( a ) ) α 1 ψ ( s ) d d s n 1 ( I a + n α , ψ f ( s ) ) | s = a + 1 Γ ( α ) a t ψ ( s ) ( ψ ( t ) ψ ( s ) ) α 1 1 ψ ( s ) d d s 1 ψ ( s ) d d s n 2 ( I a + n α , ψ f ( s ) ) d s = 1 Γ ( α + 1 ) ( ψ ( t ) ψ ( a ) ) α 1 ψ ( s ) d d s n 1 ( I a + n α , ψ f ( s ) ) | s = a + 1 Γ ( α ) ( ψ ( t ) ψ ( s ) ) α 1 1 ψ ( s ) d d s n 2 ( I a + n α , ψ f ( s ) ) | s = a s = t a t ( α 1 ) ( ψ ( t ) ψ ( s ) ) α 2 ( ψ ( s ) ) 1 ψ ( s ) d d s n 2 ( I a + n α , ψ f ( s ) ) d s = 1 Γ ( α + 1 ) ( ψ ( t ) ψ ( a ) ) α 1 ψ ( s ) d d s n 1 ( I a + n α , ψ f ( s ) ) | s = a + 1 Γ ( α ) 0 ( ψ ( t ) ψ ( a ) ) α 1 1 ψ ( s ) d d s n 2 ( I a + n α , ψ f ( s ) ) | s = a + ( α 1 ) a t ( ψ ( t ) ψ ( s ) ) α 2 ψ ( s ) 1 ψ ( s ) d d s n 2 ( I a + n α , ψ f ( s ) ) d s = 1 Γ ( α + 1 ) ( ψ ( t ) ψ ( a ) ) α 1 ψ ( s ) d d s n 1 ( I a + n α , ψ f ( s ) ) | s = a 1 Γ ( α ) ( ψ ( t ) ψ ( a ) ) α 1 1 ψ ( s ) d d s n 2 ( I a + n α , ψ f ( s ) ) | s = a + 1 Γ ( α 1 ) a t ψ ( s ) ( ψ ( t ) ψ ( s ) ) α 2 1 ψ ( s ) d d s n 2 ( I a + n α , ψ f ( s ) ) d s .
We continue with these computations, and we find
A ( t ) = B ( t ) + 1 Γ ( α n + 2 ) a t ψ ( s ) ( ψ ( t ) ψ ( s ) ) α n + 1 1 ψ ( s ) d d s 1 ( I a + n α , ψ f ( s ) ) d s ,
where
B ( t ) = 1 Γ ( α + 1 ) ( ψ ( t ) ψ ( a ) ) α 1 ψ ( s ) d d s n 1 ( I a + n α , ψ f ( s ) ) | s = a 1 Γ ( α ) ( ψ ( t ) ψ ( a ) ) α 1 1 ψ ( s ) d d s n 2 ( I a + n α , ψ f ( s ) ) | s = a 1 Γ ( α n + 3 ) ( ψ ( t ) ψ ( a ) ) α n + 2 1 ψ ( s ) d d s 1 ( I a + n α , ψ f ( s ) ) | s = a .
So we deduce
A ( t ) = B ( t ) + 1 Γ ( α n + 2 ) ( ψ ( t ) ψ ( s ) ) α n + 1 ( I a + n α , ψ f ( s ) ) | s = a s = t a t ( α n + 1 ) ( ψ ( t ) ψ ( s ) ) α n ( ψ ( s ) ) ( I a + n α , ψ f ( s ) ) d s = B ( t ) + 1 Γ ( α n + 2 ) 0 ( ψ ( t ) ψ ( a ) ) α n + 1 ( I a + n α , ψ f ( s ) ) | s = a + a t ( α n + 1 ) ( ψ ( t ) ψ ( s ) ) α n ψ ( s ) ( I a + n α , ψ f ( s ) ) d s = B ( t ) 1 Γ ( α n + 2 ) ( ψ ( t ) ψ ( a ) ) α n + 1 ( I a + n α , ψ f ( s ) ) | s = a + 1 Γ ( α n + 1 ) a t ( ψ ( t ) ψ ( s ) ) α n ψ ( s ) ( I a + n α , ψ f ( s ) ) d s = B ( t ) 1 Γ ( α n + 2 ) ( ψ ( t ) ψ ( a ) ) α n + 1 ( I a + n α , ψ f ( s ) ) | s = a + I a + α n + 1 , ψ ( I a + n α , ψ f ( t ) ) = B ( t ) 1 Γ ( α n + 2 ) ( ψ ( t ) ψ ( a ) ) α n + 1 ( I a + n α , ψ f ( s ) ) | s = a + I a + 1 , ψ f ( t ) .
Therefore, we conclude
I a + α , ψ D a + α , ψ f ( t ) = 1 ψ ( t ) d A d t ( t ) = 1 ψ ( t ) d d t 1 Γ ( α + 1 ) ( ψ ( t ) ψ ( a ) ) α 1 ψ ( s ) d d s n 1 ( I a + n α , ψ f ( s ) ) | s = a 1 Γ ( α ) ( ψ ( t ) ψ ( a ) ) α 1 1 ψ ( s ) d d s n 2 ( I a + n α , ψ f ( s ) ) | s = a 1 Γ ( α n + 3 ) ( ψ ( t ) ψ ( a ) ) α n + 2 1 ψ ( s ) d d s 1 ( I a + n α , ψ f ( s ) ) | s = a 1 Γ ( α n + 2 ) ( ψ ( t ) ψ ( a ) ) α n + 1 ( I a + n α , ψ f ( s ) ) | s = a + I a + 1 , ψ f ( t ) = α Γ ( α + 1 ) ( ψ ( t ) ψ ( a ) ) α 1 1 ψ ( s ) d d s n 1 ( I a + n α , ψ f ( s ) ) | s = a α 1 Γ ( α ) ( ψ ( t ) ψ ( a ) ) α 2 1 ψ ( s ) d d s n 2 ( I a + n α , ψ f ( s ) ) | s = a α n + 2 Γ ( α n + 3 ) ( ψ ( t ) ψ ( a ) ) α n + 1 1 ψ ( s ) d d s 1 ( I a + n α , ψ f ( s ) ) | s = a α n + 1 Γ ( α n + 2 ) ( ψ ( t ) ψ ( a ) ) α n 1 ψ ( s ) d d s 0 ( I a + n α , ψ f ( s ) ) | s = a + 1 ψ ( t ) d d t ( I a + 1 , ψ f ( t ) ) = 1 Γ ( α ) ( ψ ( t ) ψ ( a ) ) α 1 1 ψ ( s ) d d s n 1 ( I a + n α , ψ f ( s ) ) | s = a 1 Γ ( α 1 ) ( ψ ( t ) ψ ( a ) ) α 2 1 ψ ( s ) d d s n 2 ( I a + n α , ψ f ( s ) ) | s = a 1 Γ ( α n + 2 ) ( ψ ( t ) ψ ( a ) ) α n + 1 1 ψ ( s ) d d s 1 ( I a + n α , ψ f ( s ) ) | s = a 1 Γ ( α n + 1 ) ( ψ ( t ) ψ ( a ) ) α n 1 ψ ( s ) d d s 0 ( I a + n α , ψ f ( s ) ) | s = a + 1 ψ ( t ) d d t 1 Γ ( 1 ) a t ψ ( s ) ( ψ ( t ) ψ ( s ) ) 0 f ( s ) d s = j = 1 n D n j ( I a + n α , ψ f ) ( a ) Γ ( α j + 1 ) ( ψ ( t ) ψ ( a ) ) α j + 1 ψ ( t ) d d t a t ψ ( s ) f ( s ) d s = j = 1 n D n j ( I a + n α , ψ f ) ( a ) Γ ( α j + 1 ) ( ψ ( t ) ψ ( a ) ) α j + f ( t ) .
In what follows we will present the linear problem associated with our problem (1)–(2), its solution, and the corresponding Green function with its properties. We consider the fractional differential equation
D a + α , ψ u ( t ) + K ( t ) = 0 , t ( a , b ) ,
with the boundary conditions (2), where K C ( a , b ) L 1 ( a , b ) . We denote by
Δ = Γ ( α ) Γ ( α ς ) ( ψ ( b ) ψ ( a ) ) α ς 1 i = 1 m Γ ( α ) Γ ( α ϱ i ) a b ( ψ ( s ) ψ ( a ) ) α ϱ i 1 d H i ( s ) .
Lemma 6.
If Δ 0 , then the solution of problem (9)–(2) is
u ( t ) = 1 Γ ( α ) a t ψ ( s ) ( ψ ( t ) ψ ( s ) ) α 1 K ( s ) d s + ( ψ ( t ) ψ ( a ) ) α 1 Δ 1 Γ ( α ς ) a b ψ ( s ) ( ψ ( b ) ψ ( s ) ) α ς 1 K ( s ) d s i = 1 m 1 Γ ( α ϱ i ) a b a s ψ ( τ ) ( ψ ( s ) ψ ( τ ) ) α ϱ i 1 K ( τ ) d τ d H i ( s ) , t [ a , b ] .
Proof. 
By Lemma 5, the solutions of Equation (9) are
u ( t ) = I a + α , ψ K ( t ) + j = 1 n c j ( ψ ( t ) ψ ( a ) ) α j = 1 Γ ( α ) a t ψ ( s ) ( ψ ( t ) ψ ( s ) ) α 1 K ( s ) d s + j = 1 n c j ( ψ ( t ) ψ ( a ) ) α j , t [ a , b ] ,
with c i R , i = 1 , , n .
Because u ( a ) = u ( a ) = = u ( n 2 ) ( a ) = 0 , then we find c n = c n 1 = = c 2 = 0 . So Equation (12) gives us
u ( t ) = I a + α , ψ K ( t ) + c 1 ( ψ ( t ) ψ ( a ) ) α 1 = 1 Γ ( α ) a t ψ ( s ) ( ψ ( t ) ψ ( s ) ) α 1 K ( s ) d s + c 1 ( ψ ( t ) ψ ( a ) ) α 1 , t [ a , b ] .
We compute now the ψ -fractional derivatives of function u given by (13), and we obtain
D a + k , ψ u ( t ) = D a + k , ψ I a + α , ψ K ( t ) + c 1 D a + k , ψ ( ψ ( t ) ψ ( a ) ) α 1 = I a + α k , ψ K ( t ) + c 1 Γ ( α ) Γ ( α k ) ( ψ ( t ) ψ ( a ) ) α k 1 , t ( a , b ) ,
for k = ς , ϱ i , i = 1 , , m .
Therefore, the condition D a + ς , ψ u ( b ) = i = 1 m a b D a + ϱ i , ψ u ( s ) d H i ( s ) gives us
I a + α ς , ψ K ( b ) + c 1 Γ ( α ) Γ ( α ς ) ( ψ ( b ) ψ ( a ) ) α ς 1 = i = 1 m a b I a + α ϱ i , ψ K ( s ) + c 1 Γ ( α ) Γ ( α ϱ i ) ( ψ ( s ) ψ ( a ) ) α ϱ i 1 d H i ( s ) .
Then we deduce
c 1 Γ ( α ) Γ ( α ς ) ( ψ ( b ) ψ ( a ) ) α ς 1 i = 1 m Γ ( α ) Γ ( α ϱ i ) a b ( ψ ( s ) ψ ( a ) ) α ϱ i 1 d H i ( s ) = I a + α ς , ψ K ( b ) i = 1 m a b I a + α ϱ i , ψ K ( s ) d H i ( s ) ,
and so
c 1 = 1 Δ Γ ( α ς ) a b ψ ( s ) ( ψ ( b ) ψ ( s ) ) α ς 1 K ( s ) d s 1 Δ i = 1 m a b 1 Γ ( α ϱ i ) a s ψ ( τ ) ( ψ ( s ) ψ ( τ ) ) α ϱ i 1 K ( τ ) d τ d H i ( s ) ,
where Δ is given by (10). By replacing c 1 above in relation (13), we obtain the solution u of problem (9)–(2) given by (11). □
Lemma 7.
If Δ 0 , then the solution of problem (9)–(2) given by (11) can be written as
u ( t ) = a b G ( t , s ) K ( s ) d s , t [ a , b ] ,
where
G ( t , s ) = g 1 ( t , s ) + ( ψ ( t ) ψ ( a ) ) α 1 Δ i = 1 m a b g 2 i ( τ , s ) d H i ( τ ) , t , s [ a , b ] ,
and
g 1 ( t , s ) = 1 Γ ( α ) ψ ( s ) ( ψ ( t ) ψ ( a ) ) α 1 ψ ( b ) ψ ( s ) ψ ( b ) ψ ( a ) α ς 1 ( ψ ( t ) ψ ( s ) ) α 1 , a s t b , ψ ( s ) ( ψ ( t ) ψ ( a ) ) α 1 ψ ( b ) ψ ( s ) ψ ( b ) ψ ( a ) α ς 1 , a t s b , g 2 i ( τ , s ) = 1 Γ ( α ϱ i ) ψ ( s ) ( ψ ( τ ) ψ ( a ) ) α ϱ i 1 ψ ( b ) ψ ( s ) ψ ( b ) ψ ( a ) α ς 1 ( ψ ( τ ) ψ ( s ) ) α ϱ i 1 ] , a s τ b , ψ ( s ) ( ψ ( τ ) ψ ( a ) ) α ϱ i 1 ψ ( b ) ψ ( s ) ψ ( b ) ψ ( a ) α ς 1 , a τ s b ,
for i = 1 , , m .
Proof. 
By Lemma 6 and relation (11), we deduce
u ( t ) = 1 Γ ( α ) ( ψ ( b ) ψ ( a ) ) α ς 1 a t ( ψ ( t ) ψ ( a ) ) α 1 ψ ( s ) ( ψ ( b ) ψ ( s ) ) α ς 1 ( ψ ( b ) ψ ( a ) ) α ς 1 ( ψ ( t ) ψ ( s ) ) α 1 ψ ( s ) K ( s ) d s + t b ( ψ ( t ) ψ ( a ) ) α 1 ψ ( s ) ( ψ ( b ) ψ ( s ) ) α ς 1 K ( s ) d s 1 Γ ( α ) ( ψ ( b ) ψ ( a ) ) α ς 1 a b ( ψ ( t ) ψ ( a ) ) α 1 ψ ( s ) ( ψ ( b ) ψ ( s ) ) α ς 1 K ( s ) d s + ( ψ ( t ) ψ ( a ) ) α 1 Δ 1 Γ ( α ς ) a b ψ ( s ) ( ψ ( b ) ψ ( s ) ) α ς 1 K ( s ) d s i = 1 m 1 Γ ( α ϱ i ) a b a s ψ ( τ ) ( ψ ( s ) ψ ( τ ) ) α ϱ i 1 K ( τ ) d τ d H i ( s ) = a b g 1 ( t , s ) K ( s ) d s 1 Γ ( α ) × Γ ( α ) Γ ( α ς ) Δ a b ( ψ ( t ) ψ ( a ) ) α 1 ψ ( s ) ( ψ ( b ) ψ ( s ) ) α ς 1 K ( s ) d s + 1 ( ψ ( b ) ψ ( a ) ) α ς 1 × 1 Γ ( α ) Δ i = 1 m Γ ( α ) Γ ( α ϱ i ) a b ( ψ ( s ) ψ ( a ) ) α ϱ i 1 d H i ( s ) × a b ( ψ ( t ) ψ ( a ) ) α 1 ψ ( s ) ( ψ ( b ) ψ ( s ) ) α ς 1 K ( s ) d s + ( ψ ( t ) ψ ( a ) ) α 1 Δ 1 Γ ( α ς ) a b ψ ( s ) ( ψ ( b ) ψ ( s ) ) α ς 1 K ( s ) d s i = 1 m 1 Γ ( α ϱ i ) a b a s ψ ( τ ) ( ψ ( s ) ψ ( τ ) ) α ϱ i 1 K ( τ ) d τ d H i ( s ) = a b g 1 ( t , s ) K ( s ) d s + ( ψ ( t ) ψ ( a ) ) α 1 Δ 1 ( ψ ( b ) ψ ( a ) ) α ς 1 × i = 1 m 1 Γ ( α ϱ i ) a b ( ψ ( s ) ψ ( a ) ) α ϱ i 1 d H i ( s ) × a b ψ ( s ) ( ψ ( b ) ψ ( s ) ) α ς 1 K ( s ) d s i = 1 m 1 Γ ( α ϱ i ) a b a s ψ ( τ ) ( ψ ( s ) ψ ( τ ) ) α ϱ i 1 K ( τ ) d τ d H i ( s ) .
Then we obtain
u ( t ) = a b g 1 ( t , s ) K ( s ) d s + ( ψ ( t ) ψ ( a ) ) α 1 Δ × i = 1 m 1 Γ ( α ϱ i ) a b 1 ( ψ ( b ) ψ ( a ) ) α ς 1 ( ψ ( τ ) ψ ( a ) ) α ϱ i 1 d H i ( τ ) × a b ψ ( s ) ( ψ ( b ) ψ ( s ) ) α ς 1 K ( s ) d s a b τ b ( ψ ( s ) ψ ( τ ) ) α ϱ i 1 d H i ( s ) ψ ( τ ) K ( τ ) d τ = a b g 1 ( t , s ) K ( s ) d s + ( ψ ( t ) ψ ( a ) ) α 1 Δ × i = 1 m 1 Γ ( α ϱ i ) a b 1 ( ψ ( b ) ψ ( a ) ) α ς 1 ( ψ ( τ ) ψ ( a ) ) α ϱ i 1 d H i ( τ ) × a b ψ ( s ) ( ψ ( b ) ψ ( s ) ) α ς 1 K ( s ) d s a b s b ( ψ ( τ ) ψ ( s ) ) α ϱ i 1 d H i ( τ ) ψ ( s ) K ( s ) d s = a b g 1 ( t , s ) K ( s ) d s + ( ψ ( t ) ψ ( a ) ) α 1 Δ × i = 1 m 1 Γ ( α ϱ i ) a b a b 1 ( ψ ( b ) ψ ( a ) ) α ς 1 ( ψ ( τ ) ψ ( a ) ) α ϱ i 1 ψ ( s ) × ( ψ ( b ) ψ ( s ) ) α ς 1 d H i ( τ ) ) K ( s ) d s a b s b ( ψ ( τ ) ψ ( s ) ) α ϱ i 1 ψ ( s ) d H i ( τ ) K ( s ) d s = a b g 1 ( t , s ) K ( s ) d s + ( ψ ( t ) ψ ( a ) ) α 1 Δ i = 1 m a b a b g 2 i ( τ , s ) d H i ( τ ) K ( s ) d s .
Therefore, we find
u ( t ) = a b G ( t , s ) K ( s ) d s , t [ a , b ] ,
where G , g , g 2 i , i = 1 , , m are given by (15) and (16). So we obtain formula (14) for the solution u of problem (9)–(2). □
Lemma 8.
The functions g 1 , g 2 i , i = 1 , , m have the following properties for all t , s [ a , b ] :
(a) g 1 ( t , s ) h 1 ( s ) , where
h 1 ( s ) = 1 Γ ( α ) ψ ( s ) ( ψ ( b ) ψ ( s ) ) α ς 1 [ ( ψ ( b ) ψ ( a ) ) ς ( ψ ( b ) ψ ( s ) ) ς ] , s [ a , b ] .
(b) g 1 ( t , s ) ( γ ( t ) ) α 1 h 1 ( s ) , where
γ ( t ) = ψ ( t ) ψ ( a ) ψ ( b ) ψ ( a ) , t [ a , b ] .
(c) g 1 ( t , s ) ψ ( s ) Γ ( α ) ( ψ ( t ) ψ ( a ) ) α 1 .
(d) g 2 i ( t , s ) ( γ ( t ) ) α ϱ i 1 h 2 i ( s ) , where
h 2 i ( s ) = ψ ( s ) Γ ( α ϱ i ) ( ψ ( b ) ψ ( s ) ) α ς 1 ( ψ ( b ) ψ ( a ) ) ς ϱ i ( ψ ( b ) ψ ( s ) ) ς ϱ i , s [ a , b ] ,
for i = 1 , , m .
(e) g 2 i ( t , s ) ψ ( s ) Γ ( α ϱ i ) ( ψ ( t ) ψ ( a ) ) α ϱ i 1 , i = 1 , , m .
(f) g 1 , g 2 i , i = 1 , , m are continuous on [ a , b ] × [ a , b ] ; g 1 ( t , s ) 0 , g 2 i ( t , s ) 0 for all t , s [ a , b ] , g 1 ( t , s ) > 0 , g 2 i ( t , s ) > 0 , for all t , s ( a , b ) , i = 1 , , m .
Proof. 
(a) The function g 1 is nondecreasing in the first variable. Indeed, for s t , we have
g 1 t ( t , s ) = ψ ( s ) Γ ( α ) ( α 1 ) ψ ( t ) ( ψ ( t ) ψ ( a ) ) α 2 ψ ( b ) ψ ( s ) ψ ( b ) ψ ( a ) α ς 1 ( α 1 ) ψ ( t ) ( ψ ( t ) ψ ( s ) ) α 2 ] = ψ ( s ) ψ ( t ) Γ ( α 1 ) ( ψ ( t ) ψ ( a ) ) α 2 ψ ( b ) ψ ( s ) ψ ( b ) ψ ( a ) α ς 1 ( ψ ( t ) ψ ( s ) ) α 2 ψ ( s ) ψ ( t ) Γ ( α 1 ) ( ψ ( t ) ψ ( a ) ) α 2 ψ ( b ) ψ ( s ) ψ ( b ) ψ ( a ) α 2 ( ψ ( t ) ψ ( s ) ) α 2 = ψ ( s ) ψ ( t ) Γ ( α 1 ) ( ψ ( t ) ψ ( a ) ) ( ψ ( b ) ψ ( s ) ) ψ ( b ) ψ ( a ) α 2 ( ψ ( t ) ψ ( s ) ) α 2 0 .
The last inequality is true because
( ψ ( t ) ψ ( a ) ) ( ψ ( b ) ψ ( s ) ) ψ ( b ) ψ ( a ) ψ ( t ) ψ ( s ) ( ψ ( t ) ψ ( a ) ) ( ψ ( b ) ψ ( s ) ) ( ψ ( b ) ψ ( a ) ) ( ψ ( t ) ψ ( s ) ) ψ ( t ) ψ ( b ) ψ ( a ) ψ ( b ) ψ ( t ) ψ ( s ) + ψ ( a ) ψ ( s ) ψ ( b ) ψ ( t ) ψ ( a ) ψ ( t ) ψ ( b ) ψ ( s ) + ψ ( a ) ψ ( s ) ψ ( a ) ( ψ ( b ) ψ ( t ) ) + ψ ( s ) ( ψ ( t ) + ψ ( b ) ) 0 ( ψ ( b ) ψ ( t ) ) ( ψ ( s ) ψ ( a ) ) 0 , t , s [ a , b ] .
Hence g 1 ( t , s ) g 1 ( b , s ) for all t , s [ a , b ] with s t .
For s t , we obtain
g 1 t ( t , s ) = ψ ( s ) ( α 1 ) ψ ( t ) ( ψ ( t ) ψ ( a ) ) α 2 ψ ( b ) ψ ( s ) ψ ( b ) ψ ( a ) α ς 1 0 .
So g 1 ( t , s ) g 1 ( s , s ) for all t , s [ a , b ] with s t .
Therefore, we infer that g 1 ( t , s ) h 1 ( s ) for all t , s [ a , b ] , where
h 1 ( s ) = g 1 ( b , s ) = 1 Γ ( α ) ψ ( s ) ( ψ ( b ) ψ ( a ) ) α 1 ψ ( b ) ψ ( s ) ψ ( b ) ψ ( a ) α ς 1 ( ψ ( b ) ψ ( s ) ) α 1 = 1 Γ ( α ) ψ ( s ) ( ψ ( b ) ψ ( s ) ) α ς 1 ( ψ ( b ) ψ ( a ) ) ς ( ψ ( b ) ψ ( s ) ) α 1 = 1 Γ ( α ) ψ ( s ) ( ψ ( b ) ψ ( s ) ) α ς 1 ( ψ ( b ) ψ ( a ) ) ς ( ψ ( b ) ψ ( s ) ) ς 0 .
(b) For s t , by also using (18), we have
g 1 ( t , s ) = ψ ( s ) Γ ( α ) ( ψ ( t ) ψ ( a ) ) α 1 ψ ( b ) ψ ( s ) ψ ( b ) ψ ( a ) α ς 1 ( ψ ( t ) ψ ( s ) ) α 1 ψ ( s ) Γ ( α ) ( ψ ( t ) ψ ( a ) ) α 1 ψ ( b ) ψ ( s ) ψ ( b ) ψ ( a ) α ς 1 ( ψ ( t ) ψ ( a ) ) α 1 ( ψ ( b ) ψ ( s ) ) α 1 ( ψ ( b ) ψ ( a ) ) α 1 = ψ ( s ) Γ ( α ) ( ψ ( t ) ψ ( a ) ) α 1 ψ ( b ) ψ ( s ) ψ ( b ) ψ ( a ) α ς 1 ψ ( b ) ψ ( s ) ψ ( b ) ψ ( a ) α 1 = ψ ( s ) Γ ( α ) ( ψ ( t ) ψ ( a ) ) α 1 ψ ( b ) ψ ( s ) ψ ( b ) ψ ( a ) α ς 1 1 ψ ( b ) ψ ( s ) ψ ( b ) ψ ( a ) ς = ( ψ ( t ) ψ ( a ) ) α 1 ψ ( s ) ( ψ ( b ) ψ ( a ) ) α 1 Γ ( α ) ( ψ ( b ) ψ ( s ) ) α ς 1 ( ( ψ ( b ) ψ ( a ) ) ς ( ψ ( b ) ψ ( s ) ) ς ) = ψ ( t ) ψ ( a ) ψ ( b ) ψ ( a ) α 1 h 1 ( s ) = ( γ ( t ) ) α 1 h 1 ( s ) .
For s t , we find
g 1 ( t , s ) = 1 Γ ( α ) ψ ( s ) ( ψ ( t ) ψ ( a ) ) α 1 ψ ( b ) ψ ( s ) ψ ( b ) ψ ( a ) α ς 1 1 Γ ( α ) ψ ( s ) ( ψ ( t ) ψ ( a ) ) α 1 ψ ( b ) ψ ( s ) ψ ( b ) ψ ( a ) α ς 1 1 ψ ( b ) ψ ( s ) ψ ( b ) ψ ( a ) ς = ψ ( t ) ψ ( a ) ψ ( b ) ψ ( a ) α 1 1 Γ ( α ) ψ ( s ) ( ψ ( b ) ψ ( s ) ) α ς 1 ( ψ ( b ) ψ ( a ) ) ς ( ψ ( b ) ψ ( s ) ) ς = ψ ( t ) ψ ( a ) ψ ( b ) ψ ( a ) α 1 h 1 ( s ) = ( γ ( t ) ) α 1 h 1 ( s ) .
Therefore, we deduce that g 1 ( t , s ) ( γ ( t ) ) α 1 h 1 ( s ) , for all t , s [ a , b ] .
(c) For all t , s [ a , b ] , we have
g 1 ( t , s ) ψ ( s ) Γ ( α ) ( ψ ( t ) ψ ( a ) ) α 1 ψ ( b ) ψ ( s ) ψ ( b ) ψ ( a ) α ς 1 ψ ( s ) Γ ( α ) ( ψ ( t ) ψ ( a ) ) α 1 .
(d) For s t , we obtain
g 2 i ( t , s ) = 1 Γ ( α ϱ i ) ψ ( s ) ( ψ ( t ) ψ ( a ) ) α ϱ i 1 ψ ( b ) ψ ( s ) ψ ( b ) ψ ( a ) α ς 1 ( ψ ( t ) ψ ( s ) ) α ϱ i 1 ] ψ ( s ) Γ ( α ϱ i ) ( ψ ( t ) ψ ( a ) ) α ϱ i 1 ψ ( b ) ψ ( s ) ψ ( b ) ψ ( a ) α ς 1 ( ψ ( t ) ψ ( a ) ) α ϱ i 1 ( ψ ( b ) ψ ( s ) ) α ϱ i 1 ( ψ ( b ) ψ ( a ) ) α ϱ i 1 = ψ ( s ) Γ ( α ϱ i ) ( ψ ( t ) ψ ( a ) ) α ϱ i 1 ψ ( b ) ψ ( s ) ψ ( b ) ψ ( a ) α ς 1 ( ψ ( b ) ψ ( s ) ) α ϱ i 1 ( ψ ( b ) ψ ( a ) ) α ϱ i 1 = ψ ( s ) Γ ( α ϱ i ) ( ψ ( t ) ψ ( a ) ) α ϱ i 1 ψ ( b ) ψ ( s ) ψ ( b ) ψ ( a ) α ς 1 1 ψ ( b ) ψ ( s ) ψ ( b ) ψ ( a ) ς ϱ i = ψ ( s ) ( ψ ( t ) ψ ( a ) ) α ϱ i 1 Γ ( α ϱ i ) ( ψ ( b ) ψ ( a ) ) α ϱ i 1 ( ψ ( b ) ψ ( s ) ) α ς 1 × ( ψ ( b ) ψ ( a ) ) ς ϱ i ( ψ ( b ) ψ ( s ) ) ς ϱ i = ψ ( t ) ψ ( a ) ψ ( b ) ψ ( a ) α ϱ i 1 h 2 i ( s ) = ( γ ( t ) ) α ϱ i 1 h 2 i ( s ) ,
where h 2 i ( s ) is given by (17).
If t s , we infer
g 2 i ( t , s ) = 1 Γ ( α ϱ i ) ψ ( s ) ( ψ ( t ) ψ ( a ) ) α ϱ i 1 ψ ( b ) ψ ( s ) ψ ( b ) ψ ( a ) α ς 1 1 Γ ( α ϱ i ) ψ ( s ) ( ψ ( t ) ψ ( a ) ) α ϱ i 1 ψ ( b ) ψ ( s ) ψ ( b ) ψ ( a ) α ς 1 1 ψ ( b ) ψ ( s ) ψ ( b ) ψ ( a ) ς ϱ i
= ψ ( t ) ψ ( a ) ψ ( b ) ψ ( a ) α ϱ i 1 1 Γ ( α ϱ i ) ψ ( s ) ( ψ ( b ) ψ ( s ) ) α ς 1 × ( ψ ( b ) ψ ( a ) ) ς ϱ i ( ψ ( b ) ψ ( s ) ) ς ϱ i = ψ ( t ) ψ ( a ) ψ ( b ) ψ ( a ) α ϱ i 1 h 2 i ( s ) = ( γ ( t ) ) α ϱ i 1 h 2 i ( s ) .
Hence, we deduce g 2 i ( t , s ) ( γ ( t ) ) α ϱ i 1 h 2 i ( s ) , for all t , s [ a , b ] .
(e) We have
g 2 i ( t , s ) 1 Γ ( α ϱ i ) ψ ( s ) ( ψ ( t ) ψ ( a ) ) α ϱ i 1 ψ ( b ) ψ ( s ) ψ ( b ) ψ ( a ) α ς 1 ψ ( s ) Γ ( α ϱ i ) ( ψ ( t ) ψ ( a ) ) α ϱ i 1 , t , s [ a , b ] .
(f) This property follows from the definition of functions g 1 and g 2 i , i = 1 , , m , and from the properties (b) and (d) above. □
Lemma 9.
Assume that Δ > 0 and H i : [ a , b ] R , i = 1 , , m are nondecreasing functions. Then the function G given by (15) is a continuous function on [ a , b ] × [ a , b ] and satisfies the following conditions.
(a) G ( t , s ) J ( s ) , for all t , s [ a , b ] , where
J ( s ) = h 1 ( s ) + ( ψ ( b ) ψ ( a ) ) α 1 Δ i = 1 m a b g 2 i ( τ , s ) d H i ( τ ) , s [ a , b ] .
(b) G ( t , s ) ( γ ( t ) ) α 1 J ( s ) , for all t , s [ a , b ] .
(c) G ( t , s ) θ ( γ ( t ) ) α 1 , for all t , s [ a , b ] , where
θ = C 0 ( ψ ( b ) ψ ( a ) ) α 1 × 1 Γ ( α ) + 1 Δ i = 1 m a b 1 Γ ( α ϱ i ) ( ψ ( τ ) ψ ( a ) ) α ϱ i 1 d H i ( τ ) > 0 ,
with C 0 = sup s [ a , b ] ψ ( s ) .
Proof. 
By the definition of function G , we deduce that G is a continuous function. In addition, by using Lemma 8, we obtain the following for all t , s [ a , b ] :
(a)
G ( t , s ) = g 1 ( t , s ) + ( ψ ( t ) ψ ( a ) ) α 1 Δ i = 1 m a b g 2 i ( τ , s ) d H i ( τ ) h 1 ( s ) + ( ψ ( b ) ψ ( a ) ) α 1 Δ i = 1 m a b g 2 i ( τ , s ) d H i ( τ ) = J ( s ) .
(b)
G ( t , s ) ψ ( t ) ψ ( a ) ψ ( b ) ψ ( a ) α 1 h 1 ( s ) + ( ψ ( t ) ψ ( a ) ) α 1 Δ i = 1 m a b g 2 i ( τ , s ) d H i ( τ ) = ψ ( t ) ψ ( a ) ψ ( b ) ψ ( a ) α 1 h 1 ( s ) + ( ψ ( b ) ψ ( a ) ) α 1 Δ i = 1 m a b g 2 i ( τ , s ) d H i ( τ ) = ( γ ( t ) ) α 1 J ( s ) .
(c)
G ( t , s ) ψ ( s ) ( ψ ( t ) ψ ( a ) ) α 1 Γ ( α ) + ( ψ ( t ) ψ ( a ) ) α 1 Δ i = 1 m a b g 2 i ( τ , s ) d H i ( τ ) ψ ( s ) ( ψ ( t ) ψ ( a ) ) α 1 Γ ( α ) + ( ψ ( t ) ψ ( a ) ) α 1 Δ × i = 1 m a b ψ ( s ) Γ ( α ϱ i ) ( ψ ( τ ) ψ ( a ) ) α ϱ i 1 d H i ( τ ) C 0 ( ψ ( t ) ψ ( a ) ) α 1 1 Γ ( α ) + 1 Δ i = 1 m a b 1 Γ ( α ϱ i ) ( ψ ( τ ) ψ ( a ) ) α ϱ i 1 d H i ( τ ) = θ ψ ( t ) ψ ( a ) ψ ( b ) ψ ( a ) α 1 = θ ( γ ( t ) ) α 1 ,
where C 0 = sup s [ a , b ] ψ ( s ) and θ is given by (19). □
Lemma 10.
Assume that Δ > 0 , H i : [ a , b ] R , i = 1 , , m are nondecreasing functions, and K ( t ) 0 for all t ( a , b ) . Then the solution u of problem (9)–(2) given by (14) satisfies the inequality u ( t ) ( γ ( t ) ) α 1 u ( ζ ) for all t , ζ [ a , b ] .
Proof. 
By Lemma 9, we obtain the following for all t , ζ [ a , b ] :
u ( t ) = a b G ( t , s ) K ( s ) d s a b ( γ ( t ) ) α 1 J ( s ) K ( s ) d s = ( γ ( t ) ) α 1 a b J ( s ) K ( s ) d s ( γ ( t ) ) α 1 a b G ( ζ , s ) K ( s ) d s = ( γ ( t ) ) α 1 u ( ζ ) .
In Section 3 we will use the Guo–Krasnosel’skii fixed-point theorem presented below (see [49]).
Theorem 1.
Let X be a Banach space and let P X be a cone in X . Assume Y 1 and Y 2 are bounded open subsets of X with 0 Y 1 Y 1 ¯ Y 2 and let L : P ( Y 2 ¯ Y 1 ) P be a completely continuous operator such that either
(i) L u u , u P Y 1 , and L u u , u P Y 2 ;
(ii) or L u u , u P Y 1 , and L u u , u P Y 2 .
Then L has a fixed point in P ( Y 2 ¯ Y 1 ) .

3. Existence of Positive Solutions

In this section we examine the existence of positive solutions to our problem (1)–(2). Below we present the assumptions that we will use in the sequel.
(A1)
α > 0 , α ( n 1 , n ] , n N , n 3 , ψ C n ( [ a , b ] ) with ψ ( t ) > 0 for all t [ a , b ] , 1 ς < α 1 , 0 ϱ i ς for all i = 1 , , m , λ > 0 , H i : [ a , b ] R , i = 1 , , m are nondecreasing functions, and Δ > 0 (given by (10)).
(A2)
The function f C ( ( a , b ) × R + , R ) may be singular at t = a and/or t = b , and there exist the functions p , q C ( ( a , b ) , R + ) , w C ( [ a , b ] × R + , R + ) such that
q ( t ) f ( t , y ) p ( t ) w ( t , y ) , t ( a , b ) , y R + ,
with 0 < a b q ( t ) d t < and 0 < a b p ( t ) d t < , ( R + = [ 0 , ) ).
(A3)
There exist c , d ( a , b ) , c < d such that f = lim y , t [ c , d ] f ( t , y ) y = .
(A4)
There exist c ˜ , d ˜ ( a , b ) , c ˜ < d ˜ such that lim inf y min t [ c ˜ , d ˜ ] f ( t , y ) > Λ 0 , with
Λ 0 = 2 θ a b q ( s ) d s ( γ ( c ˜ ) ) α 1 c ˜ d ˜ J ( s ) d s 1 ,
and w = lim y max t [ a , b ] w ( t , y ) y = 0 , where J and θ are given in Lemma 9.
We consider the fractional differential equation
D a + α , ψ v ( t ) + λ ( f ( t , [ v ( t ) λ z ( t ) ] * ) + q ( t ) ) = 0 , t ( a , b ) ,
with the boundary conditions
v ( i ) ( a ) = 0 , i = 0 , 1 , , n 2 ; D a + ς , ψ v ( b ) = i = 1 m a b D a + ϱ i , ψ v ( s ) d H i ( s ) ,
where ϑ ( t ) * = ϑ ( t ) if ϑ ( t ) 0 , and ϑ ( t ) * = 0 if ϑ ( t ) < 0 .
Here, z ( t ) = a b G ( t , s ) q ( s ) d s , t [ a , b ] is the solution of the problem
D a + α , ψ z ( t ) + q ( t ) = 0 , t ( a , b ) , z ( i ) ( a ) = 0 , i = 0 , 1 , , n 2 ; D a + ς , ψ z ( b ) = i = 1 m a b D a + ϱ i , ψ z ( s ) d H i ( s ) .
Under assumptions ( A 1 ) and ( A 2 ) , we have z ( t ) 0 for all t [ a , b ] . We will show that there exists a solution v of problem (20)–(21), with v ( t ) λ z ( t ) on [ a , b ] , and v ( t ) > λ z ( t ) on ( a , b ) . In this case, u = v λ z represents a positive solution of problem (1)–(2). Hence, in what follows, we will study the problem (20)–(21).
By using Lemma 2, v is a solution of problem (20)–(21) if and only if v is a solution of equation
v ( t ) = λ a b G ( t , s ) ( f ( s , [ v ( s ) λ z ( s ) ] * ) + q ( s ) ) d s , t [ a , b ] .
We consider the Banach space X = C [ a , b ] with the supremum norm v = sup t [ a , b ] | v ( t ) | , and we define the cone
P = v X , v ( t ) ( γ ( t ) ) α 1 v , t [ a , b ] .
For λ > 0 , we introduce the operator
L v ( t ) = λ a b G ( t , s ) ( f ( s , [ v ( s ) λ z ( s ) ] * ) + q ( s ) ) d s ,
for t [ a , b ] , and v X .
It is clear that v is a solution of Equation (22) (or equivalently of problem (20)–(21)) if and only if v is a fixed point of operator L .
Lemma 11.
If ( A 1 ) and ( A 2 ) hold, then the operator L : P P is a completely continuous operator.
Proof. 
Let v P be fixed. By using ( A 1 ) and ( A 2 ) , we infer that L v ( t ) < . In addition, by Lemma 9, we deduce for all t , ζ [ a , b ] that
L v ( t ) λ a b J ( s ) ( f ( s , [ v ( s ) λ z ( s ) ] * ) + q ( s ) ) d s ,
and
L v ( t ) λ a b ( γ ( t ) ) α 1 J ( s ) ( f ( s , [ v ( s ) λ z ( s ) ] * ) + q ( s ) ) d s ( γ ( t ) ) α 1 L v ( ζ ) .
So L v ( t ) ( γ ( t ) ) α 1 L v for all t [ a , b ] . We conclude that L v P , and then L ( P ) P . By using standard arguments, we deduce that the operator L is completely continuous (that is, continuous, and maps bounded sets into relatively compact sets). □
Theorem 2.
Assume that ( A 1 ) , ( A 2 ) and ( A 3 ) hold. Then there exits λ 1 > 0 such that for all λ ( 0 , λ 1 ] , the boundary value problem (1)–(2) has at least one positive solution u ( t ) , t [ a , b ] .
Proof. 
We choose a positive number r 1 > θ a b q ( s ) d s , and we define the set Y 1 = { v X , v < r 1 } . We consider
λ 1 = min 1 , r 1 Ξ 1 a b J ( s ) ( p ( s ) + q ( s ) ) d s 1 ,
with Ξ 1 = max { max t [ a , b ] , y [ 0 , r 1 ] w ( t , y ) , 1 } .
Let λ ( 0 , λ 1 ] . We have
z ( t ) = a b G ( t , s ) q ( s ) d s θ ( γ ( t ) ) α 1 a b q ( s ) d s , t [ a , b ] .
Let v P Y 1 . Then for any t [ a , b ] we infer
[ v ( t ) λ z ( t ) ] * v ( t ) v = r 1 ,
and
v ( t ) λ z ( t ) ( γ ( t ) ) α 1 v λ θ ( γ ( t ) ) α 1 a b q ( s ) d s = ( γ ( t ) ) α 1 v λ θ a b q ( s ) d s ( γ ( t ) ) α 1 r 1 λ 1 θ a b q ( s ) d s ( γ ( t ) ) α 1 r 1 θ a b q ( s ) d s 0 .
Therefore we find
L v ( t ) λ a b J ( s ) ( p ( s ) w ( s , [ v ( s ) λ z ( s ) ] * ) + q ( s ) ) d s λ 1 Ξ 1 a b J ( s ) ( p ( s ) + q ( s ) ) d s r 1 = v , t [ a , b ] .
So we deduce
L v v , v P Y 1 .
Next, for c and d given by assumption ( A 3 ) , we choose a constant M 0 > 0 such that
M 0 2 λ ( γ ( c ) ) 2 ( α 1 ) c d J ( s ) d s 1 .
In addition, by ( A 3 ) , we deduce that there exists a constant M 1 > 0 such that
f ( t , y ) M 0 y , t [ c , d ] , y M 1 .
Now we define r 2 = max { 2 r 1 , 2 M 1 ( γ ( c ) ) 1 α } , and let Y 2 = { v X , v < r 2 } .
Let v P Y 2 . Then we obtain
v ( t ) λ z ( t ) ( γ ( t ) ) α 1 v λ θ ( γ ( t ) ) α 1 a b q ( s ) d s ( γ ( t ) ) α 1 r 2 θ a b q ( s ) d s ( γ ( t ) ) α 1 r 1 θ a b q ( s ) d s 0 , t [ a , b ] .
Therefore we find
[ v ( t ) λ z ( t ) ] * = v ( t ) λ z ( t ) ( γ ( t ) ) α 1 r 2 θ a b q ( s ) d s ( γ ( c ) ) α 1 r 2 2 M 1 , t [ c , d ] .
Then by (24) and (25), we deduce
L v ( t ) λ c d G ( t , s ) ( f ( s , [ v ( s ) λ z ( s ) ] * ) + q ( s ) ) d s λ c d G ( t , s ) M 0 [ v ( s ) λ z ( s ) ] * d s = λ M 0 c d G ( t , s ) ( v ( s ) λ z ( s ) ) d s λ M 0 c d ( γ ( s ) ) α 1 J ( s ) ( γ ( c ) ) α 1 r 2 2 d s λ M 0 r 2 2 ( γ ( c ) ) 2 ( α 1 ) c d J ( s ) d s r 2 = v , t [ c , d ] .
So we conclude
L v v , v P Y 2 .
Then, by (23), (26) and Theorem 1 (i), we deduce that operator L has a fixed point v 1 P ( Y 2 ¯ Y 1 ) , that is, r 1 v 1 r 2 .
Because v 1 r 1 , we find
v 1 ( t ) λ z ( t ) ( γ ( t ) ) α 1 v 1 λ θ ( γ ( t ) ) α 1 a b q ( s ) d s = ( γ ( t ) ) α 1 v 1 λ θ a b q ( s ) d s ( γ ( t ) ) α 1 r 1 θ a b q ( s ) d s = Θ 1 ( γ ( t ) ) α 1 , t [ a , b ] ,
and so v 1 ( t ) λ z ( t ) + Θ 1 ( γ ( t ) ) α 1 for all t [ a , b ] , where Θ 1 = r 1 θ a b q ( s ) d s > 0 . Let u 1 ( t ) = v 1 ( t ) λ z ( t ) for all t [ a , b ] . Then, u 1 is a positive solution of problem (1)–(2) with u 1 ( t ) Θ 1 ( γ ( t ) ) α 1 for all t [ a , b ] . □
Theorem 3.
Suppose that ( A 1 ) , ( A 2 ) , and ( A 4 ) hold. Then there exists λ 2 > 0 such that for any λ λ 2 , the boundary value problem (1)–(2) has at least one positive solution u ( t ) , t [ a , b ] .
Proof. 
By ( A 4 ) , there exists M 2 > 0 such that
f ( t , y ) Λ 0 , t [ c ˜ , d ˜ ] , y M 2 .
We define
λ 2 = M 2 ( γ ( c ˜ ) ) α 1 θ a b q ( s ) d s 1 .
Let λ λ 2 . We consider r 3 = 2 λ θ a b q ( s ) d s , and Y 3 = { v X , v r 3 } . Let v P Y 3 . Then we deduce
v ( t ) λ z ( t ) ( γ ( t ) ) α 1 v λ θ ( γ ( t ) ) α 1 a b q ( s ) d s = ( γ ( t ) ) α 1 r 3 λ θ a b q ( s ) d s = ( γ ( t ) ) α 1 λ θ a b q ( s ) d s ( γ ( t ) ) α 1 λ 2 θ a b q ( s ) d s = ( γ ( t ) ) α 1 M 2 ( γ ( c ˜ ) ) α 1 0 , t [ a , b ] .
Therefore, we find
[ v ( t ) λ z ( t ) ] * = v ( t ) λ z ( t ) ( γ ( t ) ) α 1 M 2 ( γ ( c ˜ ) ) α 1 M 2 , t [ c ˜ , d ˜ ] .
So we conclude
L v ( t ) λ c ˜ d ˜ G ( t , s ) f ( s , [ v ( s ) λ z ( s ) ] * ) d s λ c ˜ d ˜ ( γ ( t ) ) α 1 J ( s ) f ( s , v ( s ) λ z ( s ) ) d s λ ( γ ( t ) ) α 1 c ˜ d ˜ J ( s ) Λ 0 d s = λ ( γ ( t ) ) α 1 c ˜ d ˜ J ( s ) d s 2 θ a b q ( s ) d s ( γ ( c ˜ ) ) α 1 c ˜ d ˜ J ( s ) d s 1 = ( γ ( t ) ) α 1 ( γ ( c ˜ ) ) α 1 r 3 r 3 , t [ c ˜ , d ˜ ] .
Hence we obtain
L v v , v P Y 3 .
On the other hand, we consider the positive number ϵ = ( 2 λ a b J ( s ) p ( s ) d s ) 1 . Then by ( A 4 ) , we deduce that there exists M 3 > 0 such that
w ( t , y ) ϵ y , t [ a , b ] , y M 3 .
Then we find w ( t , y ) M 4 + ϵ y , for all t [ a , b ] and y 0 , where M 4 = max t [ a , b ] , y [ 0 , M 3 ] w ( t , y ) .
We now define r 4 > max { r 3 , 2 λ max { M 4 , 1 } a b J ( s ) ( p ( s ) + q ( s ) ) d s } , and Y 4 = { v X , v < r 4 } .
Let v P Y 4 . Therefore, we obtain
v ( t ) λ z ( t ) ( γ ( t ) ) α 1 v λ θ ( γ ( t ) ) α 1 a b q ( s ) d s = ( γ ( t ) ) α 1 r 4 λ θ a b q ( s ) d s ( γ ( t ) ) α 1 r 3 λ θ a b q ( s ) d s = ( γ ( t ) ) α 1 λ θ a b q ( s ) d s ( γ ( t ) ) α 1 λ 2 θ a b q ( s ) d s = M 2 ( γ ( t ) ) α 1 ( γ ( c ˜ ) ) α 1 0 , t [ a , b ] .
Then we find
L v ( t ) λ a b J ( s ) [ p ( s ) w ( s , [ v ( s ) λ z ( s ) ] * ) + q ( s ) ] d s λ a b J ( s ) [ p ( s ) ( M 4 + ϵ ( v ( s ) λ z ( s ) ) ) + q ( s ) ] d s λ max { M 4 , 1 } a b J ( s ) ( p ( s ) + q ( s ) ) d s + λ ϵ r 4 a b J ( s ) p ( s ) d s r 4 2 + r 4 2 = r 4 = v , t [ a , b ] .
Hence we obtain
L v v , v P Y 4 .
By (27), (28) and Theorem 1 (ii), we deduce that L has a fixed point v 2 P ( Y 4 ¯ Y 3 ) , so r 3 v 2 r 4 .
In addition, we find
v 2 ( t ) λ z ( t ) v 2 ( t ) λ θ ( γ ( t ) ) α 1 a b q ( s ) d s ( γ ( t ) ) α 1 v 2 λ θ ( γ ( t ) ) α 1 a b q ( s ) d s ( γ ( t ) ) α 1 r 3 λ θ ( γ ( t ) ) α 1 a b q ( s ) d s = ( γ ( t ) ) α 1 r 3 λ θ a b q ( s ) d s = ( γ ( t ) ) α 1 λ θ a b q ( s ) d s λ 2 θ ( γ ( t ) ) α 1 a b q ( s ) d s = M 2 1 ( γ ( c ˜ ) ) α 1 θ a b q ( s ) d s θ ( γ ( t ) ) α 1 a b q ( s ) d s = M 2 ( γ ( t ) ) α 1 ( γ ( c ˜ ) ) α 1 , t [ a , b ] .
We take u 2 ( t ) = v 2 ( t ) λ z ( t ) for all t [ a , b ] . Then u 2 ( t ) Θ 2 ( γ ( t ) ) α 1 for all t [ a , b ] , where Θ 2 = M 2 ( γ ( c ˜ ) ) 1 α . Therefore, we conclude that u 2 is a positive solution of problem (1)–(2). □
By using similar arguments as those from the proof of Theorem 3, we obtain the following theorem.
Theorem 4.
Assume that ( A 1 ) and ( A 2 ) hold, and
  • ( A 4 ˜ ) There exist c ˜ , d ˜ ( a , b ) , c ˜ < d ˜ such that
    f ˜ = lim y min t [ c ˜ , d ˜ ] f ( t , y ) = and w = lim y max t [ a , b ] w ( t , y ) y = 0 .
Then, there exists λ ˜ 2 > 0 such that for any λ λ ˜ 2 , the boundary value problem (1)–(2) has at least one positive solution u ( t ) , t [ a , b ] .

4. Examples

Example 1.
Let ψ ( t ) = ln t , t [ a , b ] = [ 1 , e ] , α = 7 2 , n = 4 , m = 2 , ς = 9 4 , ϱ 1 = 1 6 , ϱ 2 = 8 5 , and
H 1 ( s ) = s 4 9 5 / 3 , s 1 , 19 12 , 41 36 5 / 3 , s 19 12 , e , H 2 ( s ) = 1 , s [ 1 , 2 ) , 15 11 , s [ 2 , e ] .
We consider the Hadamard fractional differential equation
D 1 + 7 / 2 , ψ u ( t ) + λ f ( t , u ( t ) ) = 0 , t ( 1 , e ) ,
with the boundary conditions
u ( 1 ) = u ( 1 ) = u ( 1 ) = 0 , D 1 + 9 / 4 , ψ u ( e ) = 5 3 1 19 / 12 s 4 9 2 / 3 D 1 + 1 / 6 , ψ u ( s ) d s + 4 11 D 1 + 8 / 5 , ψ u ( 2 ) .
We have γ ( t ) = ln t , t [ 1 , e ] , and Δ 2.69941187 > 0 . So assumption ( A 1 ) is satisfied. In addition we find
g 1 ( t , s ) = 1 Γ ( 7 / 2 ) 1 s ( ln t ) 5 / 2 ( 1 ln s ) 1 / 4 ( ln t ln s ) 5 / 2 , 1 s t e , 1 s ( ln t ) 5 / 2 ( 1 ln s ) 1 / 4 , 1 t s e , g 21 ( τ , s ) = 1 Γ ( 10 / 3 ) 1 s ( ln τ ) 7 / 3 ( 1 ln s ) 1 / 4 ( ln τ ln s ) 7 / 3 , 1 s τ e , 1 s ( ln τ ) 7 / 3 ( 1 ln s ) 1 / 4 , 1 τ s e , g 22 ( τ , s ) = 1 Γ ( 19 / 10 ) 1 s ( ln τ ) 9 / 10 ( 1 ln s ) 1 / 4 ( ln τ ln s ) 9 / 10 , 1 s τ e , 1 s ( ln τ ) 9 / 10 ( 1 ln s ) 1 / 4 , 1 τ s e ,
and
G ( t , s ) = g 1 ( t , s ) + ( ln t ) 5 / 2 Δ 5 3 1 19 / 12 τ 4 9 2 / 3 g 21 ( τ , s ) d s + 4 11 g 22 ( 2 , s ) ,
for all t , s [ 1 , e ] . We also obtain
h 1 ( s ) = 1 s Γ ( 7 / 2 ) ( 1 ln s ) 1 / 4 ( 1 ln s ) 5 / 2 , s [ 1 , e ] , h 21 ( s ) = 1 s Γ ( 10 / 3 ) ( 1 ln s ) 1 / 4 ( 1 ln s ) 7 / 3 , s [ 1 , e ] , h 22 ( s ) = 1 s Γ ( 19 / 10 ) ( 1 ln s ) 1 / 4 ( 1 ln s ) 9 / 10 , s [ 1 , e ] .
In addition, we find θ 0.40870482 , and
J ( s ) = h 1 ( s ) + 1 Δ 1 e g 21 ( τ , s ) d H 1 ( τ ) + 1 e g 22 ( τ , s ) d H 2 ( s ) = h 1 ( s ) + 1 Δ 5 3 1 19 / 12 τ 4 9 2 / 3 g 21 ( τ , s ) d τ + 4 11 g 22 ( 2 , s ) = h 1 ( s ) + 1 Δ 5 3 1 s τ 4 9 2 / 3 1 s Γ ( 10 / 3 ) ( ln τ ) 7 / 3 ( 1 ln s ) 1 / 4 d τ + s 19 / 12 τ 4 9 2 / 3 1 s Γ ( 10 / 3 ) ( ln τ ) 7 / 3 ( 1 ln s ) 1 / 4 ( ln τ ln s ) 7 / 3 d τ + 4 11 s Γ ( 19 / 10 ) ( ln 2 ) 9 / 10 ( 1 ln s ) 1 / 4 ( ln 2 ln s ) 9 / 10 , 1 s < 19 12 , h 1 ( s ) + 1 Δ 5 3 1 19 / 12 τ 4 9 2 / 3 1 s Γ ( 10 / 3 ) ( ln τ ) 7 / 3 ( 1 ln s ) 1 / 4 d τ + 4 11 s Γ ( 19 / 10 ) ( ln 2 ) 9 / 10 ( 1 ln s ) 1 / 4 ( ln 2 ln s ) 9 / 10 , 19 12 s < 2 , h 1 ( s ) + 1 Δ 5 3 1 19 / 12 τ 4 9 2 / 3 1 s Γ ( 10 / 3 ) ( ln τ ) 7 / 3 ( 1 ln s ) 1 / 4 d τ + 4 11 s Γ ( 19 / 10 ) ( ln 2 ) 9 / 10 ( 1 ln s ) 1 / 4 , 2 s e .
We consider the function
f ( t , u ) = 3 u 2 + u + 1 ( t 1 ) 3 ( e t ) 4 2 t 1 5 , t ( 1 , e ) , u 0 .
We have q ( t ) = 2 t 1 5 , p ( t ) = 1 ( t 1 ) 3 ( e t ) 4 for all t ( 1 , e ) , w ( t , u ) = 3 u 2 + u + 1 for all t [ 1 , e ] and u 0 . We obtain 1 e q ( t ) d t 3.85492139 , and 1 e p ( t ) d t 4.44288294 . Therefore, assumption ( A 2 ) is satisfied. In addition, for c = 3 / 2 and d = 2 , the assumption ( A 3 ) is also satisfied, because f = .
After some computations using the Mathematica program, we deduce that 1 e J ( s ) ( p ( s ) + q ( s ) ) d s 0.76857453 . We choose r 1 = 2 ( > θ 1 e q ( s ) d s 1.57552 ), and we obtain Ξ 1 = 15 and λ 1 0.1735 . Then, by Theorem 2 we conclude that for any λ ( 0 , λ 1 ] , the problem (29)–(30) with the nonlinearity (31) has at least one positive solution u ( t ) , t [ 1 , e ] .
Example 2.
Let ψ ( t ) = t 3 , t [ a , b ] = [ 1 , 5 ] , α = 8 3 , n = 3 , m = 2 , ϱ 1 = 1 7 , ϱ 2 = 6 11 , and
H 1 ( s ) = 2 , s [ 1 , 3 ) , 51 25 , s [ 3 , 5 ] , H 2 ( s ) = s 1 2 1 / 41 , s [ 1 , 5 ] .
We consider the ψ-fractional differential equation
D 1 + 8 / 3 , ψ u ( t ) + λ f ( t , u ( t ) ) = 0 , t ( 1 , 5 )
with the boundary conditions
u ( 1 ) = u ( 1 ) = 0 , D 1 + 5 / 4 , ψ u ( 5 ) = 1 25 D 1 + 1 / 7 , ψ u ( 3 ) + 1 41 1 5 s 1 2 40 / 41 D 1 + 6 / 11 , ψ u ( s ) d s .
We have γ ( t ) = t 3 1 124 , t [ 1 , 5 ] , and Δ 3.61577002 > 0 . So assumption ( A 1 ) is satisfied. Besides, we find
g 1 ( t , s ) = 1 Γ ( 8 / 3 ) 3 s 2 ( t 3 1 ) 5 / 3 125 s 3 124 5 / 12 ( t 3 s 3 ) 5 / 3 , 1 s t 5 , 3 s 2 ( t 3 1 ) 5 / 3 125 s 3 124 5 / 12 , 1 t s 5 , g 21 ( τ , s ) = 1 Γ ( 53 / 21 ) 3 s 2 ( τ 3 1 ) 32 / 21 125 s 3 124 5 / 12 ( τ 3 s 3 ) 32 / 21 , 1 s τ 5 , 3 s 2 ( τ 3 1 ) 32 / 21 125 s 3 124 5 / 12 , 1 τ s 5 , g 22 ( τ , s ) = 1 Γ ( 70 / 33 ) 3 s 2 ( τ 3 1 ) 37 / 33 125 s 3 124 5 / 12 ( τ 3 s 3 ) 37 / 33 , 1 s τ 5 , 3 s 2 ( τ 3 1 ) 37 / 33 125 s 3 124 5 / 12 , 1 τ s 5 ,
and
G ( t , s ) = g 1 ( t , s ) + ( t 3 1 ) 5 / 3 Δ 1 25 g 21 ( 3 , s ) + 1 41 1 5 τ 1 2 40 / 41 g 22 ( τ , s ) d s ,
for all t , s [ 1 , 5 ] .
We also obtain
h 1 ( s ) = 1 Γ ( 8 / 3 ) 3 s 2 ( 124 ) 5 / 4 ( 125 s 3 ) 5 / 12 ( 125 s 3 ) 5 / 3 , s [ 1 , 5 ] , h 21 ( s ) = 3 s 2 Γ ( 53 / 21 ) ( 124 ) 31 / 28 ( 125 s 3 ) 5 / 12 ( 125 s 3 ) 32 / 21 , s [ 1 , 5 ] , h 22 ( s ) = 3 s 2 Γ ( 70 / 33 ) ( 124 ) 31 / 44 ( 125 s 3 ) 5 / 12 ( 125 s 3 ) 37 / 33 , s [ 1 , 5 ] .
In addition, we find θ 537632.68981746 , and
J ( s ) = h 1 ( s ) + ( 124 ) 5 / 3 Δ 1 5 g 21 ( τ , s ) d H 1 ( τ ) + 1 5 g 22 ( τ , s ) d H 2 ( τ ) = h 1 ( s ) + ( 124 ) 5 / 3 Δ 1 25 g 21 ( 3 , s ) + 1 41 1 5 τ 1 2 40 / 41 g 22 ( τ , s ) d τ
= h 1 ( s ) + ( 124 ) 5 / 3 Δ 3 s 2 25 Γ ( 53 / 21 ) ( 26 ) 32 / 21 125 s 3 124 5 / 12 ( 27 s 3 ) 32 / 21 + 1 41 Γ ( 70 / 33 ) 1 s 3 s 2 ( τ 3 1 ) 37 / 33 125 s 3 124 5 / 12 τ 1 2 40 / 41 d τ + 1 41 Γ ( 70 / 33 ) s 5 τ 1 2 40 / 41 3 s 2 ( τ 3 1 ) 37 / 33 125 s 3 124 5 / 12 ( τ 3 s 3 ) 37 / 33 d τ , 1 s < 3 , h 1 ( s ) + ( 124 ) 5 / 3 Δ 3 s 2 25 Γ ( 53 / 21 ) ( 26 ) 32 / 21 125 s 3 124 5 / 12 + 1 41 Γ ( 70 / 33 ) 1 s 3 s 2 ( τ 3 1 ) 37 / 33 125 s 3 124 5 / 12 τ 1 2 40 / 41 d τ + 1 41 Γ ( 70 / 33 ) s 5 τ 1 2 40 / 41 3 s 2 ( τ 3 1 ) 37 / 33 125 s 3 124 5 / 12 ( τ 3 s 3 ) 37 / 33 d τ , 3 s 5 .
We consider the function
f ( t , u ) = u + 1 ( t 1 ) ( 5 t ) 2 3 1 ( 5 t ) 3 4 , t ( 1 , 5 ) , u 0 .
Here we have q ( t ) = 1 ( 5 t ) 3 4 , p ( t ) = 1 ( t 1 ) ( 5 t ) 2 3 for all t ( 1 , 5 ) , w ( t , u ) = u + 1 for all t [ 1 , 5 ] and u 0 . We find 1 5 q ( t ) d t 5.65685425 and 1 5 p ( t ) d t 3.62759873 . So assumption ( A 2 ) is satisfied.
For c ˜ = 2 and d ˜ = 4 , the assumption ( A 4 ) is also satisfied, because lim inf y min t [ 2 , 4 ] f ( t , y ) = and w = 0 . After some computations by using the Mathematica program, we obtain 2 4 J ( s ) d s 230960 , and Λ 0 3170.25 . For the above Λ 0 , we find that f ( t , y ) Λ 0 for all t [ 2 , 4 ] and y M 2 , with M 2 4.5037 × 10 7 . In addition, we deduce λ 2 1783 . Therefore, by Theorem 3, we conclude that for any λ λ 2 , the problem (32)–(33) with the nonlinearity (34) has at least one positive solution u ( t ) , t [ 1 , 5 ] .

5. Conclusions

In this paper, we investigate the existence of positive solutions to the ψ –Riemann–Liouville fractional differential Equation (1), involving a positive parameter and a singular, sign-changing nonlinearity. The equation is complemented by nonlocal boundary conditions (2), which incorporate Riemann–Stieltjes integrals together with ψ –Riemann–Liouville fractional derivatives of different orders. We first construct the Green function corresponding to problem (1)–(2) and analyze its fundamental properties. Building on this, we establish our main existence results by applying the Guo–Krasnosel’skii fixed-point theorem. In the future we intend to extend the results obtained in this work to systems of ψ -fractional differential equations subject to various coupled nonlocal boundary conditions.

Author Contributions

Conceptualization, R.L.; Formal analysis, A.T. and R.L.; Methodology, A.T. and R.L.; Writing—original draft preparation, A.T. and R.L.; Writing—review and editing, A.T. and R.L. All authors have read and agreed to the published version of the manuscript.

Funding

This research received no external funding.

Data Availability Statement

The original contributions presented in this study are included in the article. Further inquiries can be directed to the corresponding author.

Acknowledgments

The authors thank the referees for their valuable suggestions and comments.

Conflicts of Interest

The authors declare no conflicts of interest.

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MDPI and ACS Style

Tudorache, A.; Luca, R. Positive Solutions for a Semipositone Singular ψ–Riemann–Liouville Fractional Boundary Value Problem. Mathematics 2025, 13, 3292. https://doi.org/10.3390/math13203292

AMA Style

Tudorache A, Luca R. Positive Solutions for a Semipositone Singular ψ–Riemann–Liouville Fractional Boundary Value Problem. Mathematics. 2025; 13(20):3292. https://doi.org/10.3390/math13203292

Chicago/Turabian Style

Tudorache, Alexandru, and Rodica Luca. 2025. "Positive Solutions for a Semipositone Singular ψ–Riemann–Liouville Fractional Boundary Value Problem" Mathematics 13, no. 20: 3292. https://doi.org/10.3390/math13203292

APA Style

Tudorache, A., & Luca, R. (2025). Positive Solutions for a Semipositone Singular ψ–Riemann–Liouville Fractional Boundary Value Problem. Mathematics, 13(20), 3292. https://doi.org/10.3390/math13203292

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